VCE Methods Question Thread!

[upandgo]

Pretty much what syndicate said, but using prime decomposition/factorization is generally a pretty fool proof method.

Here's what I mean:
(Image removed from quote.)
When you factorize a number with a factor tree, you will find every prime number that makes it up. You can look for perfect squares (4, 9, 16, 25, 36, etc.), but working all the way down usually doesn't take that much longer. From the tree, we can see that:


Or


So we can take the root of 180 by taking out the squares:

i'll definitely use this! thanks 

[she0071]

Hi! I'm not entirely sure about this question:

'A rectangle ABCD is inscribed in a circle of radius a.
Find an area-of-the-rectangle function and state the domain.'

Thanks in advance

[Adequace]

I'm not sure where to post this but are we allowed to draw symbols on our CAS calculators? Since I'm using the ClassPad, I don't think there are any shortcuts I can do with this CAS aside from just binding other symbols/commands to default keys built on the keyboard. I'll just be writing the command/symbol above the key, is this all allowed?

Didn't know how to word what I'm asking efficiently, sorry.

[she0071]

hi! There's this question I don't really understand either:
'A cuboid tank is open at the top and the internal dimensions of its base are xm and 2xm. The height is hm. The volume of the tank is Vm^3 and the volume is fixed. Let S m^2 denote the internal surface area of the tank.'

hence find S in terms of V and x

Thanks a lot in advance!

[keltingmeith]

Hi! I'm not entirely sure about this question:

'A rectangle ABCD is inscribed in a circle of radius a.
Find an area-of-the-rectangle function and state the domain.'

Thanks in advance

What aren't you sure about? How to approach it? How they got their answer? Did you possibly attempt it and get a different result, and you're not sure where you went wrong? Do you have working that you want us to possibly look through?

I'm not sure where to post this but are we allowed to draw symbols on our CAS calculators? Since I'm using the ClassPad, I don't think there are any shortcuts I can do with this CAS aside from just binding other symbols/commands to default keys built on the keyboard. I'll just be writing the command/symbol above the key, is this all allowed?

Didn't know how to word what I'm asking efficiently, sorry.

I'm not quite sure what you're asking exactly, but my understanding is that it's a question about calculator use, and not what you'll physically write on your SAC or exam?

Because in that case, you can definitely do it - you're allowed to use full functionality of your CAS as long as it doesn't result in communication with someone else.

hi! There's this question I don't really understand either:
'A cuboid tank is open at the top and the internal dimensions of its base are xm and 2xm. The height is hm. The volume of the tank is Vm^3 and the volume is fixed. Let S m^2 denote the internal surface area of the tank.'

hence find S in terms of V and x

Thanks a lot in advance!

Hmmmm, this looks [i]suspiciously[/i] like a question asked in the HSC forums. What a coincidence, eh?

[tysh]

hi! There's this question I don't really understand either:
'A cuboid tank is open at the top and the internal dimensions of its base are xm and 2xm. The height is hm. The volume of the tank is Vm^3 and the volume is fixed. Let S m^2 denote the internal surface area of the tank.'

hence find S in terms of V and x

Thanks a lot in advance!

Since V=2xh, h=V/2x. Now S = 2(x + 2h + hx) not 2(2x + 2h+ hx) since a 2x is cut from the top. Subbing in h=V/2x should give you your answer.

[upandgo]

Hi! I'm not entirely sure about this question:

'A rectangle ABCD is inscribed in a circle of radius a.
Find an area-of-the-rectangle function and state the domain.'

Thanks in advance

first step is to use pythagoras theorem:
- if we draw a diagonal line from point D to B inside the rectangle, we can label the hypotenuse as 2a (radius A multiplied by 2)
- by labelling the other two sides of the right-angle triangle as x and y, we can solve like so:

a^2 + b^2=c^2
y^2 + x^2= (2a)^2
y^2= 4a^2-x^2
hence, y= square root of 4a^2-x^2

as we are asked to find an 'area of the rectangle function', which is length multiplied by width: x(sq. root 4a^2-x^2)

and to find the domain:
                                             sq. root of 4a^2-x^2=0
x=0                                                        4a^2-x^2=0
                                                              -x^2=-4a^2
                                                                x= sq. root 4 multiplied by sq. root a^2
                                                                = 2 x a = 2a

hence, the domain is [0, 2a]

hope i helped 

[she0071]

first step is to use pythagoras theorem:
- if we draw a diagonal line from point D to B inside the rectangle, we can label the hypotenuse as 2a (radius A multiplied by 2)
- by labelling the other two sides of the right-angle triangle as x and y, we can solve like so:

a^2 + b^2=c^2
y^2 + x^2= (2a)^2
y^2= 4a^2-x^2
hence, y= square root of 4a^2-x^2

as we are asked to find an 'area of the rectangle function', which is length multiplied by width: x(sq. root 4a^2-x^2)

and to find the domain:
                                             sq. root of 4a^2-x^2=0
x=0                                                        4a^2-x^2=0
                                                              -x^2=-4a^2
                                                                x= sq. root 4 multiplied by sq. root a^2
                                                                = 2 x a = 2a

hence, the domain is [0, 2a]

hope i helped 

Yes thanks a lot! makes sense now

So for those of you keen to indulge yourself in the wonders of methods 3/4 throughout the holiday coz u have nothing better to do, post away your questions in this thread. Everyone can discuss and benefit from this, i'll personally try to answer all the questions that's posted too

the more discussion the better, once a question is posted feel free to post all kinds of solutions, talk about other things related to the question and feel free to go on mathematically related tangents, that's how one improves at maths!

[she0071]

What aren't you sure about? How to approach it? How they got their answer? Did you possibly attempt it and get a different result, and you're not sure where you went wrong? Do you have working that you want us to possibly look through?

Hmmmm, this looks [i]suspiciously[/i] like a question asked in the HSC forums. What a coincidence, eh?

yeah sorry I think I posted in the wrong section or something..

[Adequace]

Couldn't find where the 1/2 question thread went but anyway. http://m.imgur.com/a/WlAYU

For part b, in the blue is the working from my book's worked solutions. Why are you allowed to break the 11 in to 9 and 2, to then factorise them? Also doesn't it ask for one solution, but since it's a perfect square it must mean it has 2 solutions?

[cosine]

Couldn't find where the 1/2 question thread went but anyway. http://m.imgur.com/a/WlAYU

For part b, in the blue is the working from my book's worked solutions. Why are you allowed to break the 11 in to 9 and 2, to then factorise them? Also doesn't it ask for one solution, but since it's a perfect square it must mean it has 2 solutions?

Thats just a weird way of solving it, they say its the reverse of completing the square. But in this case, you can simply complete the square:






Remember, to know how many solutions it has, you can easily look at the transformations right before you work out the discriminant. Looking at this, going from the general parabolic graph, we have moved 3 units to the right, meaning one solution, but then we move 32 units up, hence there will be no solution, and to prove:





Because the discriminant is negative, there will be no solution.

[pi]

Couldn't find where the 1/2 question thread went

Mod team decided to just make this a "Methods" thread for everyone in all units just because there's so much overlap. Perhaps we should have communicated this, our apologies

[poppingsoda]

Couldn't find where the 1/2 question thread went but anyway. http://m.imgur.com/a/WlAYU

For part b, in the blue is the working from my book's worked solutions. Why are you allowed to break the 11 in to 9 and 2, to then factorise them? Also doesn't it ask for one solution, but since it's a perfect square it must mean it has 2 solutions?

Have you learnt about completing the square?

If you look at imagine the graph of the last equation, it never touches the x-axis. Therefore the discriminant is above zero for all values of m. That means that the original equation has a solution for all values of x.

[Adequace]

Thanks all, I completely forgot about completing the square here 

[keltingmeith]

Mod team decided to just make this a "Methods" thread for everyone in all units just because there's so much overlap. Perhaps we should have communicated this, our apologies

I REGRET NOTHING. /capeflick